# Question regarding kernels in Hilbert-Schmidt operators

I'm studying some Hilbert-Schmidt integral operator theory. I read that a Hilbert-Schmidt integral operator is uniquely determined by its kernel, and I wonder how exactly that is. I guess what I'm looking for are hints/suggestions on how to show that Hilbert-Schmidt integral operators are uniquely determined by their kernels.

Here are the definitions I'm going by:

$\textit{Let k be a function of two variables }(x,y)\in I \times I=I^2$ $\textit{ where }I\textit{ is a finite or infinite real interval. We define a linear integral operator K with kernel }$ $k(x,y)\textit{ as }$

$$Ku(x)=\int_{I}k(x,y)u(y)\ dy,\ \ x\in I$$

$\textit{whenever this integral makes sense. The domain D(K) will have to be specified in order to accomplish this.}$

$\textit{An integral operator on }L^2(I)\textit{ is called a Hilbert-Schmidt operator if the kernel }k\textit{ is in }L^2(I\times I),\textit{that is if}$

$$||k||_{L^{2}}^2=\int_{I}\int_{I}|k(x,y)|^2\ dxdy<\infty$$

The book I'm using now says it's possible to show that a Hilbert-Schmidt operator is uniquely determined by its kernel, that is, if

$$\int_{I}k(x,y)u(y)\ dy=\int_{I}h(x,y)u(y)\ dy$$

for all $u\in L^2(I)$, then $k=h$ in $L^2(I\times I)$.

This is where I'm unsure of how to show this. Thanks in advance!

• This should follow immediately by definition. What is the definition of a Hilbert-Schmidt integral operator that you are going by? – alw Sep 12 '18 at 19:47
• @alw I'm not really able to write the definition I'm going by since I don't have the book I'm using with me (I can post the definition later!), but the book stated that one could show the uniqueness of Hilbert-Schmidt integral operators by their kernel, which essentially is what I'm trying to do. – James Sep 12 '18 at 19:50
• @alw I have updated the question with the definitions I'm using! – James Sep 13 '18 at 16:54
• Your question is whether the kernel is uniquely determined by the Hilbert-Schmidt operator, not whether the Hilbert-Schmidt operator is uniquely determined by the kernel! – Eric Wofsey Sep 13 '18 at 20:43
• @EricWofsey That might be the case! The book said the opposite, confusingly enough. But thank you for the response and answer! – James Sep 15 '18 at 21:39

Let $T_k$ be the operator $T_ku(x)=\int_{I}k(x,y)u(y)\,dy$. If $T_k=T_h$, then $T_{k-h}=T_k-T_h=0$. So, it suffices to show that if $T_k=0$, then $k=0$.
To prove this, suppose $T_k=0$ and let $k_x(y)=\overline{k(x,y)}$. Note that $k_x\in L^2(I)$ for almost every $x\in I$, and that $T_ku(x)=\langle u, k_x\rangle$. Now pick an orthonormal basis $\{e_n\}$ for $L^2(I)$ (or really just any countable subset whose span is dense). For each $n$, $T_ke_n=0$, meaning that for almost all $x$, $k_x$ is orthogonal to $e_n$. Since a countable union of null sets is null, this means that for almost all $x$, $k_x$ is orthogonal to $e_n$ for all $n$. But since the span of $\{e_n\}$ is dense in $L^2(I)$, this means $k_x=0$ almost everywhere. So $k_x=0$ almost everywhere for almost all $x$, which means $k=0$ almost everywhere.
• Just to be sure I'm getting the proof, the bar in $\overline{k(x,y)}$ means the conjugate, right? And also, I'm not exactly sure as to why this only works for almost all $x$? I know I've come across this before but what exactly does that mean? – James Sep 15 '18 at 21:44
• Yes, the bar is complex conjugation. An element of $L^2$ is really an equivalence class of functions, where two functions are equivalent if they agree almost everywhere. Here I am dealing with specific functions representing the equivalence classes, so all my statements about functions can only be true almost everywhere. – Eric Wofsey Sep 15 '18 at 21:59
• In particular, for instance, if $k$ is some square-integrable function on $I\times I$, then $k_x$ is only guaranteed to be square-integrable for all $x$ lying outside some set of measure $0$. Similarly, the equation $T_ku(x)=\langle u,k_x\rangle$ is really an equation of elements of $L^2(I)$, so if $k$ and $u$ are actual specific functions we can only say this equation is valid for $x$ outside some set of measure $0$. – Eric Wofsey Sep 15 '18 at 22:02